Q. 34.3( 3 Votes )

# The magnetic fiel

Answer :

Given:

No. of turns/cm=50

Magnetic field inside without iron core=2.5× 10^{-3}T

Magnetic field inside solenoid with iron core=2.5T

Cross-sectional area of solenoid=4cm^{2}

We know that magnetic field(B) inside the solenoid without iron core is given by

Where,

n=no. of turns in a given length

I=current through solenoid

Putting the values of B, I and μ_{0}

A

Therefore, current through the solenoid is given by 0.4A

Magnetization I of the core is given by formula

Where H=magnetization intensity

Let the magnetic field with iron core and without iron is given by B_{2} and B_{1} respectively.

Then,

Putting the values of B_{1},B_{2} and μ

Am^{-1}

Therefore, the magnetization in the core is 2× 10^{6}A/m

Intensity of magnetization is given by the formula

Where,

M=magnetic moment given by the formula

…(i)

Where,

m=magnetic pole strength

l=length between two poles

also volume inside solenoid is given by

..(ii)

Using eqns.(i) and (ii) we get

Putting the values of I and A

A-m

A-m

Therefore, magnetic pole strength developed inside the core is given by 800A-m

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